Reusing OpenFileDialog

I have 2 textboxes and 2 button [...] next to each textbox. Is it possible to use one OpenFileDialog and pass the FilePath to the respective textbox, based on which button is clicked? i.e...if I click buttton one and laod the dialog, when I click open on the dialog, it passes the fileName to the first textbox.

From stackoverflow Saif Khan

• There are several ways to do this. One is to have a Dictionary<Button, TextBox> that holds the link between a button and its related textbox, and use that in the click event for the button (both buttons can be hooked up to the same event handler):

  public partial class TheForm : Form
  {
      private Dictionary<Button, TextBox> _buttonToTextBox = new Dictionary<Button, TextBox>();
      public Form1()
      {
          InitializeComponent();
          _buttonToTextBox.Add(button1, textBox1);
          _buttonToTextBox.Add(button2, textBox2);
      }
  
      private void Button_Click(object sender, EventArgs e)
      {
          OpenFileDialog ofd = new OpenFileDialog();
          if (ofd.ShowDialog() == DialogResult.OK)
          {
              _buttonToTextBox[sender as Button].Text = ofd.FileName;
          }
      }
  }
  

  Of course, the above code should be decorated with null-checks, nice encapsulation of behavior and so on, but you get the idea.

  From Fredrik Mörk

• Yes it is, basically you need to keep a reference to the button that was clicked, and then a mapping of a textbox to each button:

  public class MyClass
  {
    public Button ClickedButtonState { get; set; }
    public Dictionary<Button, TextBox> ButtonMapping { get; set; }
  
    public MyClass
    {
      // setup textbox/button mapping.
    } 
  
     void button1_click(object sender, MouseEventArgs e)
     {
       ClickedButtonState = (Button)sender;
       openDialog();
     }
  
     void openDialog()
     {
       TextBox current = buttonMapping[ClickedButtonState];
       // Open dialog here with current button and textbox context.
     }
  }
  

  From Russell

• This worked for me (and it's simpler than the other posts, but either of them would work as well)

  private void button1_Click(object sender, EventArgs e)
  {
      openFileDialog1.ShowDialog();
      textBox1.Text = openFileDialog1.FileName;
  }
  
  private void button2_Click(object sender, EventArgs e)
  {
      openFileDialog1.ShowDialog();
      textBox2.Text = openFileDialog1.FileName;
  }
  

  From David Stratton

• Whenever you think "there common functionality!" you should consider a method to implement it. It could look like this:

      private void openFile(TextBox box) {
          if (openFileDialog1.ShowDialog(this) == DialogResult.OK) {
              box.Text = openFileDialog1.FileName;
              box.Focus();
          }
          else {
              box.Text = "";
          }
      }
  
      private void button1_Click(object sender, EventArgs e) {
          openFile(textBox1);
      }
  

  From Hans Passant